FIRST P U C ANNUAL EXAMINATION ( MODEL PAPER-1) MARCH 2024

SUB: MATHEMATICS ( 35 )
TIME : 3 Hours 15 minutes [ Total questions : 52 ] Max. Marks : 80
Instructions : 1. The question paper has five parts namely A , B , C , D and E .
Answer all the parts .
2. Part A has 15 multiple choice questions , 5 fill in the blank questions.

PART – A
I Answer ALL multiple choice questions 15 X 1 = 1 5
1. The interval form of { x : x∈ R , -12 < x < - 10 } is
(A) [ -12 ,- 10 ) (B) [ -12 ,- 10 ] (C) ( -12 ,- 10 ] (D) ( -12 ,- 10 )
2. If ( x + 1 , y – 2 ) = ( 3 , 1 ) then the value of x and y respectively is
(A) 1 , 2 (B) 2 , 1 (C) 2 , 3 (D) 3 , 2
3. The radian measure of 120 0 is
3π 2π 6π 7π
(A) (B) (C) (D)
4 3 7 6
19 π
4. The value of tan is
3
1 1
(A) (B) (C) √3 (D) 1
√2 √3
4. If A X B = { ( 1 , 3 ) , ( 1 , 4 ) , ( 2 , 3 ) , ( 2 , 4 ) } then the set B is
(A) { 1 , 3 } (B) { 3 , 4 } (C) { 2 , 1} (D) { 2 , 4 }
5. The simplified form of (1 - i ) - ( -1 + i6) is
(A) 2 – 7 i (B) 2 + 7 i (C) 7– 2 i (D) 7+2i
6. The solution of 4x + 3 < 5x + 7 is
(A) ( -4 , ∞ ) (B) ( -2 , ∞ ) (C) ( -∞ , 4 ) (D) (- ∞ , ∞ )
7. A coin is tossed 3 times and the outcomes are recorded . Then the number of possible
outcomes are ..
(A) 3 (B) 6 (C) 8 (D) 0
8. Let Z = 3 + 2 i then the imaginary part of Z is
(A) 3 (B) 2 (C) – 2 (D) – 3
9. The third term of the sequence whose nth is given by an = (-1)n-1 5n+1 is
(A) 5 (B) 25 (C)125 (D) 625

RAJU SK GGPUC HOSAKOTE

10. The slope of the line making inclination of 60 0 with the positive direction of x -axis is
(A) 60 √3
(B) (C) √ 2 (D) 1
11. The equation of the parabola with focus ( 6 , 0 ) and directrix is x = - 6 .
(A) y 2 = 6 x (B) y 2 = 24 x (C) x2 = 24 y (D) y 2 = - 24 x
12. The octant in which the point ( 2 , 4, -7) lie .
(A) 6 (B) 7 (C) 8 (D) 5
13. The distance between the parallel lines 3x – 4 y + 7 = 0 and 3x – 4 y + 5 = 0 is
3 3 2
(A) (B) (C) (D) 0
2 5 5
14. The radius of the circle x2 + y 2 = 9 is
(A) 4 (B) 3 (C) 2 (D) 9
2
15. If is the probability of an event , then the probability of an event ‘not A’ is ,
11
2 9 1
(A) (B) 13 (C) (D)
11 11 13
II . FILL IN THE BLANKS BY CHOOSING FROM THE GIVEN BOX : 5X1=5
1 4
[ , 64 , 2 ,4 , 9 , ]
√2 3

16. A function f is defined by f(x) = 2x – 5 then the value of f(7) is -- 9

1
17. The value of sin 765 0 is -
√2
1 1 x
18. If + = then the value of x is _64
6! 7! 8!
4
19. The slope of the line 4x – 3 y – 6 = 0 is
3
sin 4 x
20. lim =2
x→0 sin 2 x
PART- B
III Answer any SIX questions 6 X 2 = 12
21. List all the elements of the following sets .
(i) A = { x : x is a letter in the word CATARACT }
(ii) B = { x : x∈N and x is a perfect cube }
Ans : (i) A = { C , A , T , R } OR A= {A,C,R ,T}
(ii) B = { 1 , 8 , 27 , 64 , ...... }

RAJU SK GGPUC HOSAKOTE

22. Let A = {3 , 6 , 9 , 12 , 15 18 , 21} and B = { 4 , 8 , 12 , 16 , 20 } find A ∩ B and A U B .
Ans : A ∩ B = { 12 }
A U B = {3 , 4 , 6 , 8 , 9 , 12 , 15 , 16 , 18 , 20 , 21}
23. A wheel makes 360 revolutions in one minute . Through how many radians does it turn in
one second ?
Ans : 360 revolutions in 1 minute means 360 revolutions in 60 seconds
Therefore , 6 revolutions in 1 second i.e, 6 X 3600 or 6X 2π = 12 π radians in 1 second .
[ Note: 1 revolution or 1 rotation = 3600 or 12 π ]
a+ib
24. If x + i y = , prove that x 2 + y 2 = 1 .
a−ib
a+ib a−ib
Ans : Given x+iy = => x-iy = ( taking conjugate )
a−ib a+ib
a+ib a−ib
( x + i y) ( x + i y ) =
a−ib a+ib
x2+y2=1. ( using ( a + b ) (a – b ) = a2 - b2 & i2 = -1 )

25. Find the multiplicative inverse of 4 – 3 i .

Ans : Let Z = 4 – 3 i , Z= 4 + 3 i |Z|2=16+ 9=25
Z̄ 4+3 i 4 3i
we know Z -1 = = > Z -1 = = +
|Z|
2
25 25 25

26. Solve 5x – 3 < 3x + 1 when (i) x is an integer (ii) x is a real number .

Ans : 5x – 3 < 3x + 1 => 2x < 4 => x < 2
(i) x is an integer x = { ........ -2 , -1 , 0 , 1 }
(ii) x is a real number . x∈ R , x∈(−∞ ,2)

27. How many chords can be drawn through 21 points on a circle ?

Ans : To draw a chord on a circle need 2 points . So the number of chords drawn through 21
n 21 21 . 20
points on a circle is C2 = C2 = = 210
2
28. Expand ( 1 – 2x )5 .
Ans : By Binomial theorem , (a + b ) 5 = a5 + 5 a4 b +10 a3 b2 + 10a b3 + 5 ab4 + b5
By putting a = 1 , b = - 2 x we get
( 1 – 2x )5 = 1 + 5 . 1. (- 2x ) + 10 . 1 .(4 x2 ) + 10 . 1. (- 8 x3 ) + 5 . 1 . 16 x4 + (-32 x5 )
( 1 – 2x )5 = 1 – 10 x + 40 x2 - 80 x3 + 80 x4 - 32 x5 .

RAJU SK GGPUC HOSAKOTE

29. Find the equation of line through ( -2 , 3 ) with slope - 4 .
Ans : we know slope-point form of equation of a line y – y1 = m ( x – x1 )
Given m = – 4 and ( x1 , y1 ) = ( – 2 , 3 )
Therefore y – 3 = – 4 ( x + 2 ) OR 4x + y + 5 = 0

(x +1)5−1
30. Evaluate : lim
x→0 x
5 5
(x +1) −1 (x+ 1) −1
Ans : lim = lim = 5 .14 = 5
x→0 x x+ 1→1 ( x +1)−1

n n
[ using lim x −a = n a n – 1 ]
x→a x−a

31. A and B are events such that P(A) = 0.42 , P(B) = 0.48 and P(A and B ) = 0.16 .
Determine (i) P( not B ) and (ii) P( A or B ) .
Ans : (i) P( not B ) = 1 - P(B) => P( not B ) = 1 – 0.48 = 0.52
(ii) P( A or B ) = P( A U B ) = P(A) + P(B) – P(A ∩ B)
P( A or B ) = 0.42 + 0.48 – 0.16 = 0.74
PART- C
IV Answer any SIX questions 6 X 3 = 18
32. Let U = { 1 , 2 , 3 , 4 , 5 , 6 } , A = { 2 , 3} and B = { 3 , 4 , 5 } .
Verify that ( A U B ) 1 = A1 ∩ B 1.
Ans : A U B = { 2 , 3 , 4 , 5 } , ( A U B ) 1 = { 1 , 6 } -----( 1)
A1 = { 1 , 4 , 5 , 6 } , B 1 = { 1 , 2 , 6 } , A1 ∩ B 1 = { 1 , 6 } -----( 2)
From (1) and (2) , we verified that ( A U B ) 1 = A1 ∩ B 1.

33. Let f( x ) = √ x and and g ( x ) = x be two functions defined over the set of
non-negative real numbers . Find ( f + g ) (x) , ( f – g ) (x) and ( f . g ) ( x ) .
Ans : Given f( x ) = √ x and and g ( x ) = x
( f + g ) (x) = f (x) + g (x) = √ x + x
( f – g ) (x) = f (x) – g (x) = √ x – x
( f . g ) (x) = f (x) . g (x) = √ x . x OR x √x OR x3/2

RAJU SK GGPUC HOSAKOTE

 3π π + 2 sec 2 π = 10 .
34. Show that 2 sin 2 + 2 cos 2
4 4 3
3π π ) = sin π = 1
Ans : we know sin = sin ( π -
4 4 4 √2
π = 1 π =2
cos , sec
4 √2 3
3π π + 2 sec 2 π
LHS = 2 sin 2 + 2 cos 2
4 4 3
1 2 1 2
=2( ) +2( ) + 2 (2)2
√2 √ 2
= 1 + 1 + 8 = 10 = RHS Hence the result .
35. Prove that cos 3 x = 4 cos 2 x – 3 cos x .
Ans : cos 3x = cos ( 2x + x ) = cos 2x . cos x – sin 2x . sin x
= ( 2 cos 2 x – 1 ) cos x – 2 sin x cos x . sin x
= 2 cos 3 x – cos x – 2 ( 1 – cos 2 x ) cos x
=2 cos 3 x – cos x – 2 cos x + 2 cos 2 x
Therefore , cos 3 x = 4 cos 2 x – 3 cos x

(3−2 i)(2+3 i)
36. Find the conjugate of
( 1+ 2i)( 2−i)
( 3−2 i)(2+3 i) 6+ 9i−4 i +6
Ans : Let Z = = [ using i 2 = –1 ]
( 1+ 2i )( 2−i) 2−i +4 i +2
12+5 i 4−3 i
Z= X [ by rationalizing ]
4 +3 i 4−3 i
48−36 i+ 20i+15 63−16i
Z= = [( a + b) (a – b) = a2 – b2 ]
16+9 25
63+16 i 63 16 i
Z- = OR Z- = +
25 25 25
x x
37. Solve : x+ + < 11
2 3
x x
Ans : x+ + < 11
2 3
multiplying 6 on both sides we get , 6x + 3x + 2x < 66
11x < 66
x<6
Therefore , For x∈ R the solution is , x∈(−∞ ,6)

RAJU SK GGPUC HOSAKOTE

38. Find the sum of first n terms and the sum of first 5 terms of the geometric series
1 + 2 + 4 + .....
3 9
2
Ans : Given a=1 ,r=
3
n
a(1−r )
The sum to n terms in G. P is , Sn = for r < 1
1−r
n
2
1(1−( ) ) n
3
Sn = = 3 (1−( 2 ) )
2 3
1−
3
5
211
For n = 5 , S5 = 3 (1−( 2 ) ) OR S5 =
3 81
39. Find the angles between lines √ 3 x + y = 1 and x + √ 3 y = 1 .
Ans : m1 = - √3
1
m2 = -
1
√3
we know tan θ = |m1−m2
1+m1 . m2 | [ slope = -
a
b
]

| | ||
1 −2
−√ 3+
tan θ =
1+1
√3 = √3
2
= |−1
√ 3|
1 π
θ = tan -1 = OR θ = 30 0
√3 6
40. Find the equation of the ellipse whose vertices are ( ± 5 , 0 ) and foci are ( ± 4 , 0 ) .
Ans : Given vertices ( ± a , 0 ) = ( ± 5 , 0 ) and focii = ( ± c , 0 ) = ( ± 4 , 0 )
Therefore , a=5 , c=4
For ellipse, c2 = a2 – b2 => 16 = 25 – b 2 => b 2 = 9 => b = 3
2 2 2 2
Required equation of ellipse is x 2 + y2 =1 => x y
+ =1
a b 25 9

41. Are the points A ( 3 , 6 , 9) , B ( 10 , 20, 30 ) and C ( 25 , -41 , 5 ) are the vertices of a
right angled triangle ?
Ans : Using distance formula AB = √( x −x ) +( y − y ) +( z −z )
2 1
2
2 1
2
2 1
2

AB = √ 49+196+121 = √ 366 BC = √ 225+3721+ 625 = √ 4571

AC = √ 484+ 2209+ 16 = √ 2709
Now AB2 + AC2 = 366 + 2709 = 3075 and BC 2 = 4571
since AB2 + AC2 ≠ BC 2 Therefore , given points are not the vertices of a
right angled triangle .

RAJU SK GGPUC HOSAKOTE

42. Compute the derivative of cos x from first principle method .
Ans : Let f(x) = cos x and f( x + h ) = cos ( x + h )
dy f ( x+ h)−f (x )
we know = lim
dx h→0 h
x +h+ x x +h−x
[−2 cos( ).sin ( )]
dy cos ( x+ h)−cos( x ) 2 2
= lim = lim
dx h→0 h h→0 h
h
[. sin( )]
dy x+ h+ x 2
=- lim cos ( ). lim
dx h→0 2 h→0 h /2
d (cos x)
= - cos x
dx
PART- D
Answer any FOUR questions 4 x 5 = 20
43. Define greatest integer function. Draw its graph . Write the domain and range .
Ans : The function f : R → R defined by f(x) = [ x ] , x∈R assumes the value of the greatest
integer , less than or equal to x . Such a function is called the greatest integer function.
Ex: [ x ] = - 1 for – 1 ≤ x < 0

Domain = R
Range = Z ( integers )

cos 4 x+ cos 3 x+ cos 2 x

44. Prove that = cot 3x
sin 4 x+sin 3 x +sin 2 x
cos 4 x+ cos 3 x+ cos 2 x
Ans : LHS =
sin 4 x+sin 3 x +sin 2 x
( cos 4 x +cos 2 x)+cos 3 x
=
(sin 4 x +sin 2 x )+ sin 3 x
(2 cos 3 x .cos x )+cos 3 x C +D C−D
= [ using cos C+cos D=2cos( )sin ( )]
( 2sin 3 x . cos 2 x)+sin 3 x 2 2
cos 3 x .(cos x+ 1)
=
sin 3 x (cos x+ 1)
= cot 3x = RHS

RAJU SK GGPUC HOSAKOTE

45. Find the number of arrangements of the letters of the word INDEPENDENCE .
In how many of these arrangement
a) do all the vowels always occur together ? b) do all the vowels never occur together.
Ans : The number of arrangements of the word INDEPENDENCE
12!
is = 1663200
3 !2 !4 !
8! 5!
a) All vowels occur together is = . = 16800
3!2! 4!
5!
[ taking IEEEE as 1 letter ( this can be arranged in ) and others 7 letters
4!
8!
1+ 7 = 8 letters can be arranged in ]
3 !2 !
b) All the vowels never occur together = All arrangements – all vowels occur together
= 1663200 – 16800 = 1646400

46. Prove that for every positive integer n

(a + b ) n = nC0 an + nC1 an-1 b+ nC2 an -2 b2 + ---------+ n
Cn −1 a bn-1 + nCn b n
Ans : we can prove this by using Mathematical Induction
Step – 1 : For n = 1 , ( a + b )1 = a + b
The result is true for n = 1 .
Step – 2 : Assume the result is true for n = k
i.e , (a + b )k = ak + k
C1 ak-1 b+ ---------+
k
C k−1 a bk-1 + b k ------(i)
Step – 3 : To prove the result is true for n = k + 1
multiplying (a + b ) on both sides of equation (i) we get ,
(a + b )k+1 = (a + b ) [ak + k
C1 ak-1 b+ ---------+ k
C k−1 a bk-1 + b k ]
= a [ak + k
C1 ak-1 b+ ---------+
k
C k−1 a bk-1 + b k ]
+ b [ak + k
C1 ak-1 b+ ---------+
k
C k−1 a bk-1 + b k ]
= ak +1 + k
C1 ak b+ ---------+
k
C k−1 a2 bk-1 + a b k
+ bak + k
C1 ak-1 b2+ ---------+
k
C k−1 a bk + b k+1
Therefore , (a + b )k+1 = ak+1 + k+1
C 1 ak+1-1 b+ ---------+
k+1
C k+1−1 a bk+1-1 + b k +1
[ using k
C1 + k
C0 = k+1
C1 and k
C2 + k
C1 = k+1
C2 and so on ]
Hence the result is true for n = k + 1
Step – 4 : Therefore , P(n) is true for all natural number n .

RAJU SK GGPUC HOSAKOTE

47. Derive the distance of a point ( x1 , y1 ) from the line A x + B y + C = 0 .
Ans : Given the line A x + B y + C = 0 [ to reduce to normal form ]
divide by √ A 2+ B2 , we get P( x1 , y1)
A B −C d
x+ y= -----(i) M
√ A +B
2 2
√ A +B
2 2
√ A +B
2 2

This is of the form x cos ɑ + y sin ɑ = p O ←Ax+By+C= 0

−C
where p =
√ A2 + B 2
A line parallel to (i) and passing through ( x1 , y1 ) is
A B
x1 + y1 = p1
√ A +B
2 2
√ A +B2 2

From fig the distance of a point ( x1 , y1 ) from the line A x + B y + C = 0 is

A B −C
d= PM=OP- OM=p1 – p = x1 + y1 -
√ A +B2 2
√ A +B
2 2
√ A2 + B 2
Ax 1+ By 1+C
d=| | [ as distance is positive ]
√ A 2 +B 2
tan x
48. Prove that lim =1 where x is in radian measure .
x→0 x
Ans : A c
Let O be the radius and r be the radius and x be the angle
at the centre . B
O x M
From fig ,
Area of Δ OAB < Area of sector OAB < Area of Δ OBC
1 1 2 1
x OB x AM < r x< x OB x BC
2 2 2
1 1 2 1
r . r sin x < r x< r. r tan x From Δ OAM,
2 2 2
1 2
divide r , we get AM = r sin x
2
sin x < x < tan x From Δ OBC,
x
Divide tan x , cos x < < 1
tan x
1 tan x
taking reciprocals , > >1
cos x x
tan x
taking limit as x → 0 and by sand witch theorem , we get lim =1
x→0 x

RAJU SK GGPUC HOSAKOTE

49. Find the mean deviation about the mean for the following data .
Marks obtained 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Number of 2 3 8 14 8 3 2
students
Ans :
M.O xi fi fi xi |xi - x| fi| xi-x|
10-20 15 2 30 30 60
20-30 25 3 75 20 60
30-40 35 8 280 10 80
40-50 45 14 630 0 0
50-60 55 8 440 10 80
60-70 65 3 195 20 60
70-80 75 2 150 30 60
Σfi=40 1800 400
Σf i x i 1800
N = Σ fi = 40 , X= = = 45
N 40
Σf i|xi − x̄|
M.D ( mean ) = => M . D ( x ) = = 400 = 10
N 40

50. A bag contains 9 discs of which 4 are red , 3 are blue and 2 are yellow . The discs are
similar in shape and size . A disc is drawn at random from the bag . Calculate the
probability that it will be (i) red (ii) yellow (iii) not blue (iv) either blue or red .
Ans : Given A bag contains 9 discs of which 4 are red , 3 are blue and 2 are yellow .
4
(i) P( red ) =
9

(ii) P( yellow ) = 2
9
3 6 2
(iii) P( not blue ) = 1 – P(blue) = 1 - = =
9 9 3

( iv) P( either blue or red) = P(blue) + P( red ) = 3 + 4 = 7

9 9 9
[ Note : The event of selecting blue or red ball is mutually exclusive ]

RAJU SK GGPUC HOSAKOTE

 PART-E
Answer the following questions
51. Prove geometrically that cos ( x+ y ) = cos x cos y – sin x sin y 6
OR
2 2
Define hyperbola and derive its equation in the standard form x 2 − y2 =1 , a > b .
a b
Ans :

RAJU SK GGPUC HOSAKOTE

52. Find the sum of the sequence 5 , 55 , 555 , 5555 , ........... to n terms . 4
OR
x +cos x
Find the derivative of With respect to x .
tan x
Ans : Let Sn = 5 + 55 + 555 + 5555 + ........... to n terms
= 5 [ 1 + 11 + 111 + 1111 + ......... to n terms ]
5
= [ 9 + 99 + 999 + 9999 + ......... to n terms ]
9
5
= [ ( 10 – 1 ) + ( 100 – 1 ) + (1000 – 1) + (10000 – 1) + ....... to n terms ]
9
5
= [ 10 + 102 + 103 + 104 + ................– ( 1 + 1 + 1 + ........to n terms ) ]
9
n
a(r −1)
Here a = 10 , r = 10 in G. P with Sn = and 1+1+1+....to n terms = n
r−1
5 10 (10n−1)
Sn = [ -n]
9 9

Let u = x + cos x u1 = 1 – sin x

v = tan x v1 = sec 2 x
1 1 1
u vu −uv
( ) = 2
v v
x +cos x 1 tan x(1−sin x)−(x+ cos x)sec 2 x
( ) = 2
tan x tan x

RAJU SK GGPUC HOSAKOTE

 FIRST P U C ANNUAL EXAMINATION (MODEL PAPER -2 ) FOR MARCH 2024
SUB: MATHEMATICS ( 35 )
TIME : 3 Hours 15 minutes [ Total questions : 52 ] Max. Marks : 80
Instructions : 1. The question paper has five parts namely A , B , C , D and E .
Answer all the parts
2. Part A has 15 multiple choice questions , 5 fill in the blank questions.

PART – A
I Answer ALL multiple choice questions 15 X 1 = 1 5
1. The following is an example for an infinite set.
(A) { x : x∈N and x2 = 4 } (B) { x : x∈N and x is a prime }
(C) { x : x∈N and 2x - 1 = 0 } (D) { x : x∈ N and (x – 1 ) ( x – 2 ) = 0}
2. If A X B = { ( p , q ) , ( p , r ) , ( m , q ) , ( m , r ) } then the set B is
(A) { p , q } (B) { m , r } (C) { q , r } (D) { m , q }
3. The radian measure of 210 0 is
3π 2π 6π 7π
(A) (B) (C) (D)
4 3 7 6
31 π
4. The value of sin is
3

(A)
1
(B)
1
(C) √3 (D) 1
2 √2 2
5. The simplified form of 3(7 + i 7 ) + i ( 7 + i7) is
(A) 14 – 28 i (B) 7 + 14 i (C) 49 - 7 i (D) 14 + 28 i
6. The solution of 4x + 3 < 6x + 7 is
(A) ( -2 , 4 ) (B) ( -2 , ∞ ) (C) ( -∞ , 4 ) (D) (- ∞ , ∞ )
7. If n
C9 =
n
C8 then the value of n
C17 is
(A) -1 (B) 1 (C) 17 (D) -17
8. The number of terms in the expansion of ( a+ b ) n is
(A) n (B) n–1 (C) n + 1 (D) 3
9. The fourth term whose n th term is given by a n = n (n + 2) is
(A) 24 (B) 10 (C) n + 2 (D) 15
10. The equation of a line with slope 2 and y intercept is -3 is
(A) y = 2 x –3 (B) y = 3 x – 2 (C) x = 2 y – 3 (D) y = 2 x

RAJU SK GGPUC HOSAKOTE

11. The equation of the parabola with focus ( 3 , 0 ) and directrix is x = - 3 is
(A) y 2 = 12 x (B) x2 = 12 y (C) x2 = 3 y (D) y 2 = 3 x
12. The octant in which the point ( -2 , 4, -7) lie .
(A) 6 (B) 7 (C) 8 (D) 5
13. lim x ( x +1) =
x→3

(A) 6 (B) 4 (C) 12 (D) 3

14. The mean of the data 4 , 7 , 8 , 9 , 10 , 12 , 13 , 9 .
(A) 8 (B) 9 (C) 7 (D) 10
15. Given P(A) = 3 and P(B) = 1 , if A and B are mutually exclusive events
5 5
then P(A or B ) is
2 4 3 1
(A) (B) (C) (D)
5 5 25 5
II . FILL IN THE BLANKS BY CHOOSING FROM THE GIVEN BOX : 5X1=5
3 3
[ , 4 , 28 , 0 , , 16 ]
2 5
16. The number of relations from A= { 1 , 2 } to B = { 3 , 4 } is ---16

17. The value of cos ( - 1710 0 ) is -- 0

8!
18. The value of is --28
6 !2 !

19. The distance of the point ( 3 , - 5 ) from the line 3 x – 4 y - 26 = 0 is _ 3

5
x 15−1 3
20. lim 10
= 2
x→1 x −1
PART- B
III Answer any SIX questions 6 X 2 = 12
21. List all the elements of the following sets .
(i) A = { x : x is an odd natural number }
(ii) B = { x : x is a month of a year not having 31 days }
Ans : (i) A = { 1 , 3 , 5 , 7 , ......... }
(ii) B = { February , April , June , September , November }

RAJU SK GGPUC HOSAKOTE

22. Let A = {1 , 2 , 3 , 4} and B = { 3 , 4 , 5 , 6 } find A ∩ B and A U B .
Ans : A ∩ B = { 3 , 4 }
A U B = {1 , 2 , 3 , 4 , 5 , 6 }
23. In a circle of diameter 40 cm , the length of a chord is 20 cm . Find the length of minor
arc of the chord .
Ans : Given d= 40 cm , radius = 20 cm Chord = 20 cm
o
Hence θ = π as it is a equilateral triangle 2
3
20 π
We know s = rθ , Therefore s = 20 . π = s
3 3

24. Express 5+ √ 2 i in the form of a+ib.

1−√ 2 i

Ans : Let Z = 5+ √ 2 i By rationalizing

1−√ 2 i

Z = 5+ √2 i X 1+ √2 i
1−√ 2 i 1+ √2 i

Z = 5+ 5 √2 i+ √2 i−2 [ using (a + b ) ( a – b ) = a2 – b2 & i2 = -1 ]

1+ 2
3+ 6 √ 2 i
Z= =1+2 √2 i
3
25. Find the multiplicative inverse of 2 – 3 i .
Ans : Let Z=2–3i , Z = 2 + 3 i and |Z|2 = 4 + 9 = 13
Z̄ 2+3 i 2 3
we know Z -1 = Therefore , Z -1 = = + i
|Z|2 13 13 13
26. Solve 30 x < 200 when (i) x is a natural number , (ii) x is an integer.
200 20
Ans : Given 30 x < 200 => x < => x < =6.666
30 3
(i) x is a natural number => x = { 1 , 2 , 3 , 4 , 5 , 6 }
(ii) x is an integer => x = { ............. -1 , 0 , 1 , 2 , 3 , 4 , 5 , 6 }
n
P4 5
27. Find the value of n such that n −1
= ,n>4.
P4 3
n
P4 5 n( n−1)(n−2)(n−3) 5
Ans : Given n −1
= => =
P4 3 (n−1)(n−2)(n−3)(n−4) 3
n 5
= => 5n – 3n = 20 => n = 10
n−4 3

RAJU SK GGPUC HOSAKOTE

 3 4
28. Expand ( x2 + ) , x ≠0
x
Ans : By Binomial theorem , (a + b ) 4 = a4 + 4 a3 b +6 a2 b2 + 4a b3 + b4
3
By putting a = x2 , b= we get
x
3 4 3 3 2 3 3 3 4
( x2 + ) = (x2)4 + 4 . (x2)3 . ( ) +6 .( x2 )2 ( ) + 4 (x2 ) ( )+( )
x x x x x
3 4 108 81
( x2 + ) = x8 + 12 x5 + 54 x2 + +
x x x4
29. Find the equation of the line parallel to the line 3 x – 4 y + 2 = 0 passing through
the point ( - 2 , 3 ) .
Ans : A line parallel to the line 3 x – 4 y + 2 = 0 is of the form 3 x – 4 y + k = 0
This passes through the point ( - 2 , 3 ) ,i.e, 3 (- 2 ) – 4 (3) + k = 0
=> k = 18
Therefore , 3 x – 4 y + 18 = 0 be the required line .
3 2
x −4 x + 4 x
30. Evaluate : lim
x→2 x2 −4
2 2
x 3−4 x 2 + 4 x x (x −4 x +4 ) x(x −2)
Ans : lim = lim = lim [ ]
x→2 x2 −4 x→2
2
x −4 x→2 ( x−2)(x +2)
x (x−2) 2( 0)
= lim [ ] = =0
x→2 (x+ 2) 4
1 1 1
31. If E and F are events such that P(E) = , P(F) = and P(E and F ) = , find
4 2 8
(i) P( E or F ) (ii) P( not E and not F ) .
Ans : (i) P( E or F ) = P( E U F )
= P(E) + P(F) – P(E ∩ F)
1 1 1
= + -
4 2 8
5
P( E U F ) =
8
(ii) P( not E and not F ) = P( E1 ∩ F1 )
= P( E U F )1
5
= 1 - P( E U F ) = 1 -
8
3
P( not E and not F ) =
8

RAJU SK GGPUC HOSAKOTE

 PART- C
IV Answer any SIX questions 6 X 3 = 18
32. Let U = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 } , A = { 2 , 4 , 6 , 8 } and B = { 2 , 3 , 5 , 7 } .
Verify that ( A U B ) 1 = A1 ∩ B 1.
Ans : A U B = { 2 , 3 , 4 , 5 , 6 , 7 , 8} , ( A U B ) 1 = { 1 , 9 } -----( 1)
A1 = { 1 , 3 , 5 , 7 , 9 } , B 1 = { 1 , 4 , 6 , 8 , 9 } , A1 ∩ B 1 = { 1 , 9 } -----( 2)
From (1) and (2) , we verified that ( A U B ) 1 = A1 ∩ B 1.
33. Let f ( x ) = x 2 and and g ( x ) = 2 x + 1 be two real functions
then find ( f + g ) (x) , ( f – g ) (x) and ( f . g ) ( x ) .
Ans : Given f( x ) = x 2 and and g ( x ) = 2x + 1
( f + g ) (x) = f (x) + g (x) = x 2+ 2x + 1
( f – g ) (x) = f (x) – g (x) = x 2– (2x + 1) = x 2– 2x – 1
( f . g ) (x) = f (x) . g (x) = x 2 (2x + 1) OR 2x3 + x2
34. Show that tan 3 x tan 2 x tan x = tan 3 x – tan 2 x – tan x.
Ans : tan 3x = tan ( 2x + x )
tan 2 x + tan x tan x+ tan y
tan 3x = [using tan (x + y)= ]
1−tan 2 x tan x 1−tan x tan y
tan 3x [ 1 – tan 2x tan x ] = tan 2x + tan x
tan 3x – tan 3x tan 2x tan x = tan 2x + tan x
tan 3x – tan 2x – tan x = tan 3x tan 2x tan x Hence the result .
35. Prove that cos 4 x = 1 - 8 sin 2 x cos 2 x .
Ans : cos 4x = cos 2 (2x)
= 1 – 2( sin 2x )2 [ using cos 2x = 1- 2 sin2 x ]
= 1- 2 ( 2 sin x cos x )2 [ using sin 2x = 2 sin x cos x]
cos 4x = 1 - 8 sin 2 x cos 2 x
( 3−2 i)(2+3 i)
36. Find the conjugate of .
( 1+ 2i )( 2−i)
( 3−2 i)(2+3 i) 6+ 9i−4 i +6
Ans : Let Z = = [ using i 2 = –1 ]
( 1+ 2i )( 2−i) 2−i +4 i +2
12+5 i 4−3 i
Z= X [ by rationalizing ]
4 +3 i 4−3 i
48−36 i+ 20i+15 63−16i
Z= = [( a + b) (a – b) = a2 – b2 ]
16+9 25

Z- = 63+16 i OR Z- = 63 + 16 i
25 25 25

RAJU SK GGPUC HOSAKOTE

 x
37. Solve : > x +1.
3 2
x x
Ans : Given > +1 multiplying 6 on both sides , we get
3 2
2x > 3x + 6
-x>6 OR x < - 6
Hence , for x∈ R the solution is , x∈(−∞ ,−6)
38. A person has 2 parents , 4 grandparents , 8 great grandparents , and so on . Find the
number of his ancestors during the ten generations preceding his own .
Ans : Here a = 2 , r = 2 n = 10
n
a(r −1)
The sum to n terms in G. P is , Sn = for r > 1
r−1
10
2( 2 −1)
S10 = = 2(1024-1) = 2046
2−1
Hence , 2046 are his ancestors during the ten generations preceding his own .
39. Derive slope - intercept form of equation of a line .
Ans : Let m be the slope the slope and c be the
y- intercept and p( x , y ) be any point on the line .
y2 − y1
Then slope =
x2 −x1
y −c
m=
x−0
Therefore , y = m x + c be the slope - intercept form of equation of a line .
40. Find the coordinates of the focus , axis, the equation of the directrix and latus rectum
of the parabola y 2 = 12 x .
Ans : Given y 2 = 4 . 3 x => a=3
The coordinates of the focus = ( a , 0) = ( 3 , 0)
Axis is x – axis
the equation of the directrix is x = - a i.e, x = - 3
and length of latus rectum = 4a = 12
41. Are the points A ( 0 , 7 , -10) , B ( 1 , 6 , -6 ) and C ( 4 , 9 , -6 ) are the vertices of an
isosceles triangle ?
Ans :Using distance formula AB = √(x 2−x 1)2 +( y 2 − y 1)2 +( z 2−z 1)2
AB = √ 1+1+16 = √ 18 BC = √ 9+9+0 = √ 18
AC = √ 16+4 +16 = √ 36
Now AB2 + BC2 = 18 + 18 = 36 and AC 2 = 36
since AB2 + BC2 = AC 2
Therefore , given points are the vertices of a right angled triangle .

RAJU SK GGPUC HOSAKOTE

42. Compute the derivative of log x from first principle method .
Ans : Let f(x) = log x and f( x + h ) = log ( x + h )
dy f ( x+ h)−f (x )
we know = lim
dx h→0 h
( x +h)
[log ]
dy log ( x +h)−log ( x) x
= lim = lim
dx h→0 h h→0 h
x 1
dy 1 h
= lim log (1+ ) h ¿ [ lim (1+ x) x ¿=e ]
dx x h→0 x x→0

d (log x) 1
= [ loge e = 1 ]
dx x
PART- D
V Answer any FOUR questions 4 x 5 = 20
43. Define modulus function. Draw its graph . Write its domain and range .

Ans : A function f : R →R defined by f( x ) = { - x if x< 0

{ x if x ≥ 0 is called modulus function.

Domain = R (Real numbers ) and Range = R + ( positive real numbers )

sin 5 x−2 sin3 x +sin x

44. Prove that = tan x .
cos 5 x −cos x
sin 5 x−2 sin3 x +sin x
Ans : LHS =
cos 5 x −cos x
sin 5 x+ sin x−2 sin 3 x
=
cos 5 x −cos x
2 sin 3 x cos 2 x−2sin 3 x
=
−2 sin3 x sin 2 x
−2 sin 3 x (1−cos 2 x )
= [Remember 1- cos 2x = 2 sin2 x]
−2sin 3 x sin2 x
2
(1−cos 2 x) 2 sin x
= = [Remember sin 2x = 2 sin x cos x ]
sin 2 x 2 sin x cos x
sin x
= = tan x = RHS
cos x

RAJU SK GGPUC HOSAKOTE

45. Find the number of arrangements of the letters of the word INDEPENDENCE .
In how many of these arrangement
a) do the words start with P? b) do the words begin with I and end in P ?
Ans : The number of arrangements of the word INDEPENDENCE (12 letters )
12!
is = 1663200 [ N- 3 , D-2 , E-4 ]
3 !2 !4 !
a) Now the words start with p means fix p in the beginning and arrange remaining
11 !
is = = 138600
3 !2 ! 4 !
b) similarly the words begin with I and end in P means fix I in the first and p in the last
10!
remaining 10 letters have to arranged is =12600
3 !2 ! 4 !

46. Prove that for every positive integer n

(a + b ) n = n
C0 an +
n
C1 an-1 b+
n
C2 an -2 b2 + ---------+
n
Cn −1 a bn-1 + nCn b n
Ans : we can prove this by using Mathematical Induction
Step – 1 : For n = 1 , ( a + b )1 = a + b
The result is true for n = 1 .
Step – 2 : Assume the result is true for n = k
i.e , (a + b )k = ak + k
C1 ak-1 b+ ---------+
k
C k−1 a bk-1 + b k ------(i)
Step – 3 : To prove the result is true for n = k + 1
multiplying (a + b ) on both sides of equation (i) we get ,
(a + b )k+1 = (a + b ) [ak + k
C1 ak-1 b+ ---------+ k
C k−1 a bk-1 + b k ]
= a [ak + k
C1 ak-1 b+ ---------+
k
C k−1 a bk-1 + b k ]
+ b [ak + k
C1 ak-1 b+ ---------+
k
C k−1 a bk-1 + b k ]
= ak +1 + k
C1 ak b+ ---------+
k
C k−1 a2 bk-1 + a b k
+ bak + k
C1 ak-1 b2+ ---------+
k
C k−1 a bk + b k+1
Therefore , (a + b )k+1 = ak+1 + k+1
C 1 ak+1-1 b+ ---------+
k+1
C k+1−1 a bk+1-1 + b k +1
[ using k
C1 +
k
C0 =
k+1
C1 and k
C2 +
k
C1 =
k+1
C2 and so on ]
Hence the result is true for n = k + 1
Step – 4 : Therefore , P(n) is true for all natural number n .

RAJU SK GGPUC HOSAKOTE

47. Derive the distance of a point P( x1 , y1 ) from a line Ax + By + C = 0 .
Ans : Given the line A x + B y + C = 0 [ to reduce to normal form ]
divide by √ A 2+ B2 , we get P( x1 , y1)
A B −C d
x+ y= -----(i) M
√ A +B
2 2
√ A +B
2 2
√ A +B
2 2

This is of the form x cos ɑ + y sin ɑ = p O ←Ax+By+C= 0

−C
where p =
√ A2 + B 2
A line parallel to (i) and passing through ( x1 , y1 ) is
A B
x1 + y1 = p1
√ A +B
2 2
√ A +B2 2

From fig the distance of a point ( x1 , y1 ) from the line A x + B y + C = 0 is

A B −C
d= PM=OP- OM=p1 – p = x1 + y1 -
√ A +B 2 2
√ A +B
2 2
√ A2 + B 2
Ax 1+ By 1+C
d=| | [ as distance is positive ]
√ A 2 +B 2
sin x
48 . Prove that lim =1 where x is in radian measure .
x→0 x
Ans : A c
Let O be the radius and r be the radius and x be the angle
at the centre . B
O x M
From fig ,
Area of Δ OAB < Area of sector OAB < Area of Δ OBC
1 1 2 1
x OB x AM < r x< x OB x BC
2 2 2
1 1 2 1
r . r sin x < r x< r. r tan x From Δ OAM,
2 2 2
1 2
divide r , we get AM = r sin x
2
sin x < x < tan x From Δ OBC,
x 1
Divide sin x , 1 < < BC = r tan x
sin x cos x
sin x
taking reciprocals , 1 > > cos x [ cos 0 = 1 ]
x
sin x
taking limit as x → 0 and by sand witch theorem , we get lim =1
x→0 x

RAJU SK GGPUC HOSAKOTE

49. Find the mean deviation about the median for the following data .
xi 5 7 9 10 12 15

fi 8 6 2 2 2 6

xi fi cf | xi - M| fi| xi - M|

5 8 8 2 16
7 6 14 0 0
9 2 16 2 4
10 2 18 3 6
12 2 20 5 10
15 6 26 10 60
Σfi= 26 96

13 th term +14 thterm 7 +7

median = = = 7
2 2
Σf i|xi −M|
M.D ( median => M . D (M) = 96 = 3.69
N 26
50. A committee of two persons is selected from two men and two women . What is the
probability that the committee will have (a) no man ? (b) one man (c) two men ?
Ans : Here selection of persons and so order is not important .
2 men + 2 women = 4 persons
4X3
A committee of two persons is selected from 4 persons is 4C 2 = =6.
2
(a) The probability that the committee will have no man is
2 2
C0 X C2 1X1 1
P( no man ) = 4 = =
C2 6 6
(b) The probability that the committee will have one man is
2 2
C1 X C1 2X2 4 2
P( one man ) = 4
= = =
C2 6 6 3
(c) The probability that the committee will have two men is
2 2
C2 X C0 1X1 1
P( two men ) = 4 = =
C2 6 6

RAJU SK GGPUC HOSAKOTE

 PART-E
VI Answer the following questions
51. Prove geometrically that cos ( x+ y ) = cos x cos y – sin x sin y 6
OR
2 2
Define ellipse and derive its equation in the standard form x 2 + y2 =1 , a > b .
a b

RAJU SK GGPUC HOSAKOTE

52. Find the sum of the sequence 7 , 77 , 777 , 7777 , ........... to n terms . 4
OR
x 5−cos x
Find the derivative of With respect to x .
sin x
Let Sn = 7 + 77 + 777 + 7777 + ........... to n terms
= 7 [ 1 + 11 + 111 + 1111 + ......... to n terms ]
7
= [ 9 + 99 + 999 + 9999 + ......... to n terms ]
9
7
= [ ( 10 – 1 ) + ( 100 – 1 ) + (1000 – 1) + (10000 – 1) + ....... to n terms ]
9
7
= [ 10 + 102 + 103 + 104 + ................– ( 1 + 1 + 1 + ........to n terms ) ]
9
n
a(r −1)
Here a = 10 , r = 10 in G. P with Sn = and 1+1+1+....to n terms = n
r−1
n
7 10 (10 −1)
Sn = [ -n]
9 9

Let u = x5 - cos x u1 = 5 x4 + sin x

v = sin x v1 = cos x
1 1 1
u vu −uv
( ) = 2
v v
5
x −cos x
1
sin x (5 x 4 +sin x)−( x 5−cos x )cos x
( ) =
sin x sin2 x
4 2 5 2
cos x +cos x
= 5 x sin x+ sin x−x
2
sin x
4 5
dy
= 5 x sin x−x2 cos x +1
dx sin x

RAJU SK GGPUC HOSAKOTE